In the given figure- - ABC and - OBC are both isosceles-It can be concluded that-

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SSS Criteria for Congruency

In the given ...

Question

$In the given figure, Δ A B C and Δ O B C are both isosceles.$
$It can be concluded that:$

$Δ A O B ≅ Δ A O C$

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AB = AO

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$∠ A B O = ∠ A C O$

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$Δ A O B is not congruent to Δ A O C$

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Solution

The correct option is C
$∠ A B O = ∠ A C O$

$In Δ A O B and Δ A O C, A B = A C (Sides of isosceles Δ A B C) . O B = O C (Sides of isosceles Δ O B C) . A O i s c o m m o n ∴ Δ A O B ≅ Δ A O C (by SSS congruence)$
And $∠ A B O = ∠ A C O$ ( By CPCT)

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In the given figure, ΔABC and ΔOBC are both isosceles. It can be concluded that:

The correct option is C ∠ABO=∠ACO In ΔAOB and ΔAOC,AB=AC(Sides of isosceles ΔABC).OB=OC(Sides of isosceles ΔOBC).AO is common∴ΔAOB≅ΔAOC(by SSS congruence) And ∠ABO=∠ACO ( By CPCT)

$In the given figure, Δ A B C and Δ O B C are both isosceles.$
$It can be concluded that:$

The correct option is C
$∠ A B O = ∠ A C O$

$In Δ A O B and Δ A O C, A B = A C (Sides of isosceles Δ A B C) . O B = O C (Sides of isosceles Δ O B C) . A O i s c o m m o n ∴ Δ A O B ≅ Δ A O C (by SSS congruence)$
And $∠ A B O = ∠ A C O$ ( By CPCT)